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3 years ago

A 2000 N net force will give a car with some amount of mass an acceleration of 4 m/s2

Physics

1 answer:

hammer [34]3 years ago

3 0

Answer:

2m/s²

Explanation:

Given parameters:

Net force on the car = 2000N

Acceleration = 4m/s²

Unknown:

Acceleration of a car twice the mass = ?

Solution:

Let us first find the mass of the car;

Force = mass x acceleration

Mass = $\frac{Force }{acceleration}$

Mass = $\frac{2000}{4}$ = 500kg

Now,

whose mass is twice that of the car

Mass of the new car = 2 x 500 = 1000kg

So;

Acceleration = $\frac{Net force }{mass}$

Acceleration = $\frac{2000}{1000}$ = 2m/s²

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The air temperature in a 28 in.3 container with a free sliding piston is initially measured at 45 °

sergiy2304 [10]

To solve this we assume that the gas inside is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = 659.7 x 28 / 504.7

V2 = 36.60 in^3

8 0

3 years ago

What is the value of work done on an object when a 70-newton force moves it 9.0 meters in the same direction as the force?

marta [7]

630 watts - 9 x 70 = 630

4 0

4 years ago

Read 2 more answers

How far does a car travel in 90 seconds if it’s traveling 55 m/s? Show equation

podryga [215]

You just said the car is traveling at the speed of 55 m/s. If I understand this correctly, that means the car will cover:

55 meters in the first second,

55 meters in the 2nd second,

55 meters in the 3rd second,

55 meters in the 4th second,

55 meters in the 5th second,

55 meters in the 87th second,

55 meters in the 88th second,

55 meters in the 89th second, and

55 meters in the 90th second.

That's 55 meters 90 times. If you just move these words around a little bit, it says "90 times 55 meters" . . . a pretty simple arithmetic problem.

The equation is . . . Distance = (55 m/s) times (time, in seconds) .

I get 4,953 meters. You should check me on this.

8 0

3 years ago

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

user100 [1]

Answer:

A) $\Delta P = 14512.5 Pa = 14.512 kPa$

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as $v_2 = 150 m/s$

since inside air is atmospheric , $v_1 = 0 m/s$

a) By using bernoulli equation between outside and inside of flight

$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh$

$\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$

$\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]$

$\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]$

$\Delta P = 14512.5 Pa = 14.512 kPa$

b) force exerted on window

Area of window $= 25\times 45 = 1125 cm^2 = 0.1125 m^2$

We know that force is given as

$F = P\times A$

$F = 14512.5 \times 0.1125$

F = 1632.65 N

5 0

4 years ago

4. Describe how the velocity of an object changes if it undergoes uniformly acceleration motion. Can its direction change?

valentinak56 [21]

Answer:

n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

Explanation:

Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.

In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

$a_{t}$ = (v₂-v₁)/Δt

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

$a_{c}$ = v2/R

In the general case, both the module and the address change

a = Ra ( a_{t}^2 + a_{c}^2)

4 0

3 years ago